Find $x_1,x_2$ such that $\min_{x_1,x_2} x_1^2+2x_1x_2$ where $x_1,x_2$ are subject to constraint $x_1^2x_2 \ge 10$.

Question

I am attempting to solve a constrained optimization problem using Lagrange multipliers but am getting lost on how to resolve the equations the gradients output.

The problem is the following:

Find $x_1,x_2$ such that $\min_{x_1,x_2} x_1^2+2x_1x_2$ where $x_1,x_2$ are subject to constraint $x_1^2x_2 \ge 10$.

I have changed the constraint into the equality $x_1^2x_2-10-s^2=0$ and attained the gradients which result in 4 equations and 4 unknowns:

\begin{align} x^2_1x_2-10-s^2 &= 0 \\ 2x_1+2x_2 &= \lambda (2x_1x_2) \\ 2x_1 &= \lambda x_1^2 \\ 0 &= \lambda(-2s) \end{align}

But I am unsure of how to proceed from here. Additionally, I am struggling to find the dual problem.

Answer 1

The fourth equation implies that $\lambda=0$ or $s=0$. Suppose $\lambda=0$. Then the third equation implies that $x_1=0$, which contradicts the original constraint. So $s=0$. Because $x_1 \not= 0$, the third equation yields $x_1 = 2 / \lambda$, which reduces the second equation to $$4/\lambda + 2x_2 = 4x_2,$$ which yields $x_2 = 2/\lambda = x_1$. Now the first equation becomes $x_1^3 = 10$, which means that $x_1=x_2=\sqrt[3]{10}$.

Answer 2

Clearly $x_1 \neq 0$ since $x_1^2x_2 \ge 10$. Thus $2 = \lambda x_1\implies 2x_1+2x_2=4x_2\implies x_1=x_2$. Since $\lambda \neq 0 \implies s = 0 \implies x_1^3 = 10\implies x_1 = \sqrt[3]{10} = x_2$. Thus $\text{min}(x_1^2+2x_1x_2)= 3x_1^2=3\sqrt[3]{100}$.

Alternative method:

$x_2 \ge \dfrac{10}{x_1^2}\implies x_1^2+2x_1x_2 \ge x_1^2+\dfrac{20}{x_1}=f(x_1)$. By using first derivative, you can get the same answer as the Lagrange Multiplier method.